Subject: Chemistry — Chemical Equilibrium
Question: The plot of $\log k_{f}$ versus $\frac{1}{T}$ for a
reversible reaction $A(g) \rightleftharpoons P(g)$ is shown.
[IMAGE] Pre-exponential factors for the forward and backward reactions are
$10^{15}s^{- 1}$ and $10^{11}s^{- 1}$, respectively.
If the value of $\log K$ for the reaction at $500\ K$ is 6, the value
of $\left| \log k_{b} \right|$ at $250\ K$ is \_\_\_\_\_\_\_
\[K = equilibrium constant of the reaction
$k_{f} =$ rate constant of forward reaction
$k_{b} =$ rate constant of backward reaction\]
Answer: (5)
Solution: $\frac{1}{T} = 0.002$
$$T = 500$$
$$\log k_{f} = 9$$
$$k_{f} = 10^{9}$$
$$\log K_{eq} = 6$$
$$K_{eq} = 10^{6} = \frac{k_{f}}{k_{b}} = \frac{10^{9}}{k_{b}}$$
$k_{b} = 10^{3}$ (at 500 K)
$$A_{b} = 10^{11}$$
$$\log k_{b} = \log A_{b} - \frac{(Ea)_{b}}{2.303R(500)}$$
$$3 = 11 - \frac{(Ea)_{b}}{2.303R(500)}$$
$$\frac{(Ea)_{b}}{2.303R} = 4000$$
$\left| \log k_{b} \right|$ at 250 ?
$$\log k_{b} = \log A_{b} - \frac{(Ea)_{b}}{2.303R(250)}$$
$$= 11 - \frac{4000}{250} = 11 - 16$$
$$\left| \log k_{b} \right| = 5$$